Chapter 3

Thin Films

Contents3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.2 Basics of interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.3 Transfer Matrix Formulation for Multilayer Systems . . . . . . . . . . . . . . . . . 3–103.4 Applications of thin films . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–17

3.1 Introduction

This chapter deals with the propagation of optical waves in and through thin films. Thin filmsare used for a very wide range of optical applications ranging from a thin metal film for use as amirror to stacks of quarter-wave layers for highly reflective coatings in laser cavities. Thin filmsalso serve as material systems for integrated photonic circuits such as polymer films for polymerwaveguide circuits or SOI (silicon-on-insulator) circuits. In this chapter we will gain insight in theoptical properties of thin films and provide tools to deal with thin film problems in a quantitativeway.

The problem considered is schematically represented in Figure 3.1: a light source generates electro-magnetic waves and illuminates a stack of layers of different media. It is clear that the propagationof electromagnetic waves generated by the light source will be heavily influenced by the layeredstack of media. Along their path the travelling light waves will be reflected and transmitted mul-tiple times by interfaces between different media. As a result, different waves will contribute tothe electromagnetic field in each layer of the stack giving rise to interference phenomena. As ageneral result, part of the light incident on the stack of layers will be transmitted, part of it will beabsorbed and part of it will be reflected. In this chapter we will introduce a systematic approachthat allows to calculate the reflection and transmission of light at such a stack of layers. Moreover,this approach can be used to calculate the electromagnetic field in all layers of the stack, even for avery high number of layers. This method uses the transfer matrices for wave propagation throughlayers and at interfaces and will provide an efficient toolbox for a wide variety of thin film opticalproblems. Throughout this chapter, we consider linear, homogeneous and isotropic media. How-

3–1

Figure 3.1: The problem of light propagation through layered media.

ever, we do not restrict ourselves to lossless media. As we will see, the transfer matrix formalismof wave propagation perfectly allows to deal with complex refractive indices.

Consider a general light source producing light with a discrete or continuous spectrum of fre-quencies. Given that the materials involved are linear, each frequency component of the lightinteracts independently with the media. Therefore it is legitimate to investigate the properties oflayered media using monochromatic light, that is light with a unique frequency of oscillation, anddo so for every frequency component in the spectrum. Moreover, we learned in the chapter onFourier optics that in any plane in space each vector component of the electric (or magnetic) fieldU(x, y, z) of a monochromatic wave can be expanded into plane waves. It follows that we canlimit our investigation to the propagation of monochromatic plane waves.

The phenomenon of interference is fundamental for the understanding of the optical behaviour ofthin films. This follows from the insight that multiple reflections and refractions by the differentinterfaces in the stack give rise to numerous field contributions that interfere with each other.Therefore, we will first treat interference before we go on with the transfer matrix formalism. Wewill end by investigating some examples of thin film applications.

3.2 Basics of interference

The term interference generally indicates that two (or more) phenomena are interacting with eachother. Within the context of electromagnetism, this term may be a bit misleading. The reason isthis: wave propagation of electromagnetic fields is described by Maxwell’s equations. In linearmedia, these equations are linear differential equations and thus, waves will not interact with eachother: the total field vector is always the sum of the individual field vectors. In other words, for awave problem with two (or more) sources, the propagating fields do not influence each other, asthe principle of superposition holds.

However, propagating waves often seem to interact in a non-linear way. This is because the totalfield depends strongly on the phase difference between the waves. In the end, this means that thetotal intensity is (often) not equal to the sum (or superposition) of the individual intensities. Forexample, if two waves with the same frequency and the same intensity come together, the totalintensity can take any value between zero (destructive interference) and four times (constructiveinterference) the value of the intensity of the individual wave, depending on their relative phase

3–2

difference. The reason for the apparent interaction is that the quantity observed is not the ampli-tude of the waves but the energy density, which is a non-linear function of the electromagneticfield. Note that the concept of phase difference only makes sense for sources with (nearly) equalwavelengths.

3.2.1 Intensity

Any detector - be it the human eye, a photographic plate or an optical power meter - will detectthe energy density U(r, t) associated with the total electric field E(r, t). The response time of adetector is of course finite and therefore, a time-averaged energy density 〈U(r, t)〉will be detected,defined as the intensity of the wave:

I = I(r, t) = 〈U(r, t)〉 = ε 〈E(r, t) · E(r, t)〉 , (3.1)

where the time-average of the function U(r, t) is defined as

〈U(r, t)〉 =1T

∫ T

0U(r, t)dt. (3.2)

The averaging is performed over a time interval T.

If two fields E1(r, t) and E2(r, t) are present, the time-averaged energy density is the result of thetotal field E(r, t) which is the superposition of E1(r, t) and E2(r, t):

I = I(r, t) = 〈U(r, t)〉= ε 〈(E1(r, t) + E2(r, t)) · (E1(r, t) + E2(r, t))〉= ε 〈E1(r, t) · E1(r, t) + E2(r, t) · E2(r, t) + 2E1(r, t) · E2(r, t)〉 .= I1 + I2 + 2I12

(3.3)

The first two terms in 3.3 are the time-averaged energy densities or intensities associated witheach wave separately. The third term is a cross-term that gives rise to interference phenomena.The same holds when multiple waves interfere. In that case, multiple cross terms will add to thesum of the individual intensities. Before we start a more detailed discussion of the cross-term, wefirst investigate the aspect of polarization within the context of interference.

3.2.2 Polarization

The vectorial nature of the cross-term is of great importance. The cross-term is a dot-product oftwo vectors and is zero only when the two vectors have orthogonal directions. For the case ofplane waves, this implies that interference occurs always except when the waves have orthogonalpolarization states. This property will help us to simplify the mathematical treatment of interfer-ence problems. Consider for example a plane wave, propagating in the z-direction:

E(z, t) = <[Ae+j(ωt−kz)]. (3.4)

3–3

A is a complex vector that lies in the xy-plane and is defined as

A = Axejφx(t)ex +Aye

jφy(t)ey, (3.5)

or equivalently, the x and y-components of the field vector E are described by:

Ex = Ax cos(ωt− kz + φx(t))Ey = Ay cos(ωt− kz + φy(t)),

(3.6)

with Ax and Ay positive numbers.

As we know from chapter 2, the phase difference δ(t) = φx(t)− φy(t) determines the polarizationstate (linear, circular, elliptic, partial, unpolarized) of the plane wave. Each of the components ExandEy can be regarded as linearly polarized plane waves with orthogonal directions of oscillation.And given that those waves will never interfere, they can be treated separately. In conclusion, anyinterference problem of waves with arbitrary polarization states can be treated by first decom-posing the waves into orthogonally polarized wave components and then investigate interferencebetween the wave components of colinear polarization (thus having parallel field vectors). This iswhat we will do in the next section.

3.2.3 Interference of two plane waves

Consider two monochromatic plane waves with frequencies ω1 and ω2, wave vectors k1 and k2

and with colinear polarisation.

E1(r, t) = A1 cos(ω1t− k1 · r + φ1(t))E2(r, t) = A2 cos(ω2t− k2 · r + φ2(t)).

(3.7)

Since both fields have colinear polarization, we may discard the vectorial nature of the fields andwrite for each of the fields (n = 1, 2):

En(r, t) = An cos(ωnt− kn · r + φn(t)). (3.8)

The intensity associated with this plane wave can be easily calculated and yields:

I = ε⟨|An|2 cos2 (ωnt− kn · r + φn(t))

⟩= ε

2 |An|2 〈1 + cos (2 [ωnt− kn · r + φ(t)])〉

= ε2 |An|

2

(3.9)

One will get the same result when working with the complex vector notation: the intensity isdirectly related to the modulus squared of the complex amplitude.

We will now examine the effect of the cross term in detail. As we will be calculating energy-densities involving products of field vectors, it is safe practice to use the real field vector notation.

The intensity or time-averaged energy density thus becomes:

3–4

I = ε⟨E2

1

⟩+ ε⟨E2

2

⟩+ 2ε 〈E1E2〉

= εA2

1

2+ ε

A22

2+ εA1A2 [〈cos [(ω1 − ω2)t− (k1 − k2) · r + φ1(t)− φ2(t)]〉] +

εA1A2 [〈cos [(ω1 + ω2)t− (k1 + k2) · r + φ1(t) + φ2(t)]〉]= I1 + I2 + 2

√I1I2 [〈cos [(ω1 − ω2)t− (k1 − k2) · r + φ1(t)− φ2(t)]〉] +

2√I1I2 [〈cos [(ω1 + ω2)t− (k1 + k2) · r + φ1(t) + φ2(t)]〉] ,

(3.10)

where I1 and I2 are the intensities of the individual waves.

The sum frequency term 〈cos [(ω1 + ω2)t− (k1 + k2) · r + φ1(t) + φ2(t)]〉 will be averaged out byevery detector as T >> 1

ω1+ω2, so we find:

I = I1 + I2 + 2√I1I2 〈cos [(ω1 − ω2)t− (k1 − k2) · r + φ1(t)− φ2(t)]〉 . (3.11)

If ω1 6= ω2, the cross term will be non-zero only when 1ω1−ω2

>> T , i.e. if the two frequencies areonly slightly different. The result is a time-dependent beating of 〈U〉. However, when 1

ω1−ω2<< T

the last term in 3.11 will be zero.

If ω1 = ω2, then the phase difference between the two fields is K · r + δ where K = k2 − k1 andδ(t) = φ1(t) − φ2(t). If δ(t) is constant over a long enough period of time (>> T ), then the twowaves are mutually coherent and a stationary interference pattern will be observed in space. ForA1 = A2 = A0 and thus I1 = I2 = I0 the interference pattern is described by:

I = 2εA20

2 [1 + 〈cos(K · r + δ(t))〉]= 4I0 cos2(K·r+δ(t)

2 )= 4I0 cos2(Φ

2 ),(3.12)

where Φ = K · r + δ(t) and I0 is the intensity of the individual waves. The intensity thus variesperiodically in space in the direction of K and varies between 0 and 4 times the intensity of a singlewave. The spatial period of the interference pattern is

∆ =2π|K|

=λ0

2sin(θ/2), (3.13)

where θ is the angle between the two wave vectors and λ0 = 2πc/ω is the vacuum wavelength oflight.

Equation (3.12) indicates a strong dependence of the total intensity on the phase difference. Thisdependence is plotted in Figure 3.3. Also when the two waves originate from the same source, butpropagate along different paths with different optical path lengths before they come together, asillustrated in figure 3.2, the detected intensity will depend on the value of the phase Φ = K·r+δ(t).

Assuming a path length difference (also called delay) of d, the phase equals

Φ =2πdλ

=2πndλ0

(3.14)

3–5

Figure 3.2: Interference of two waves originating from the same source, after having travelled a differentpath length

Figure 3.3: Interference of two waves, (a) I1 = I2 = I0, (b) I1 6= I2.

An interferometer uses the strong phase-dependence of the intensity to measure small variationsof distance d, index n or wavelength λ0 (or frequency ν). If d/λ = 104, then an index variationof δn = 10−4 realizes a phase difference δφ = 2π. Analogously, the phase changes over 2π, if dincreases with a wavelength δd = λ. An increase of the frequency δν = c/d has the same effect. Atiny change in optical path length causes a change in Φ and thus a change in detected intensity.

3.2.4 Interference of multiple plane waves

The optical properties of thin films are mostly due to interference effects of more than two waves.When multiple waves come together, the total field is given by

E(r, t) =N∑n=1

En(r, t) (3.15)

and the intensity of the total field can be calculated by definition (3.1)

Considering fields of colinear polarization,

En(r, t) = An cos(ωnt− kn · r + φn(t)), (3.16)

we may discard the vectorial nature of the fields to calculate the intensity of the total field:

3–6

I =N∑n=1

In +N∑n=1

N∑m=1,m 6=n

√(In)

√(Im) 〈cos [(ωn − ωm)t− (kn − km) · r + φn(t)− φm(t)]〉 , (3.17)

where In = ε |An|22 . The first term is the sum of the individual intensities and the second term

determines the interference pattern.

In this section, we will briefly examine two special cases that are regularly encountered in thinfilm applications. To simplify the notations and calculations, we will from now on work with thecomplex notation of the field amplitudes.

Interference of M plane waves with equal amplitudes and constant phase difference

Consider M plane waves Em = Amejωt with complex amplitudes defined by

Am =√I0e

[j(m−1)δ], m = 1, 2, . . . ,M. (3.18)

The waves have equal intensity I0 and a constant phase difference δ. We define h = ejδ, so thatAm = I0

1/2hm−1. The complex amplitude of the total wave becomes

A =√I0

(1 + h+ h2 + . . .+ hM−1

)=

√I0

1− hM

1− h

=√I0

1− ejMδ

1− ejδ(3.19)

and the intensity is

I = |A|2 = I0

∣∣∣∣∣e−jMδ/2 − ejMδ/2

e−jδ/2 − ejδ/2

∣∣∣∣∣2

(3.20)

so that

I(δ) = I0sin2 (Mδ/2)sin2 (δ/2.)

(3.21)

This function is plotted in Figure 3.4 for different values of M. For M=2, the same interferencepattern is found as described above. For higher values of M, the interference pattern exhibitspeaks. Indeed, for δ → 0, we find that

limδ/2→0

I(δ) = I0M2, (3.22)

3–7

Figure 3.4: Interference of a finite number of plane waves with equal amplitudes and constant phase dif-ference: intensity I(δ)

M2I0in function of phase difference δ for (a) M = 2, (b) M = 4, (c) M = 8 and (d)

M = 16.

indicating constructive interference of all the fields. This happens each time when δ = 2mπ.However, for certain values of δ, namely 2π

M , 4πM , ..., 2(M−1)π

M , I(δ) becomes 0. That is why theinterference patterns exhibit M − 1 minima and M − 2 secondary maxima.

This example of interference between M waves is common in practice. Probably the most well-known case is the illumination of a screen through M slits by a plane wave. The diffracted fieldexhibits the behavior described above, in function of the angle.

Interference of an infinite number of waves with progressively declining amplitude and equalphase difference

Let us now consider an infinite number of plane waves with exponentially decreasing amplitudecoming together. The complex amplitudes are given by:

A1 =√I0, A2 = hA1, A3 = hA2 = h2A1, . . . (3.23)

with now h = |h| ejδ and |h| < 1. I0 is again the intensity of the initial wave. The superposition ofall these waves has complex amplitude

A = A1 +A2 +A3 + . . .

=√I0(1 + h+ h2 + . . .)

=√I0

1− h

=√I0

1− |h| ejδ, (3.24)

which can be rewritten as

3–8

Figure 3.5: Interference of an infinite number of waves with progressively declining amplitude and equalphase difference: intensity I in function of phase difference δ.

I =I0

(1− |h|)2 + 4 |h| sin2(δ/2). (3.25)

or

I =Imax

1 +(

2Fπ

)2sin2

(δ2

) (3.26)

withImax =

I0

(1− |h|)2(3.27)

and

F =π |h|

12

1− |h|(3.28)

a parameter called finesse.

As illustrated in figure 3.5 the intensity is a periodic function of δ with period 2π. It reaches themaximum Imax for δ = 2πq, with q an integer. When the finesse F is large (so r is close to one),the function I is sharply peaked. As the finesse F decreases the peaks become less sharp and theydisappear when r = 0. An important parameter associated with this interference pattern, is theso-called Full Width at Half Maximum (FWHM), equal to

∆δ =2πF. (3.29)

The finesse F is the ratio between the period 2π of the peaks and the FWHM of the transmissionpeaks.

This example is especially relevant in practice, in particular for the Fabry-Perot interferometer. Wewill come back to this structure in section 3.4.1.

3–9

Figure 3.6: Reflection and transmission at a multilayer stack.

3.3 Transfer Matrix Formulation for Multilayer Systems

We will now investigate the general problem of wave propagation in and through a multilayerstructure. Consider the dielectric structure depicted in Figure 3.6. The multilayer structure can beregarded as an optical system with one input and one output port. Thus, a transfer matrix T canbe assigned to the system which relates the incident and reflected waves at the input port with theincident and reflected waves at the output port. If light is incident from the left, the reflected andtransmitted field can thus be calculated once the transfer matrix T is known. In this section wewill build the matrix T by regarding the multilayer stack as a cascaded system of interfaces andlayers, each with their own transfer matrix. We get the complete transfer matrix T by multiplyingthe individual transfer matrices.

The layered medium consists of a stack of layers with thickness di and refractive indices ni, sepa-rated by interface planes. A plane wave with wave vector k is incident on the stack, as depicted inFigure 3.6. It is convenient to choose the following coordinate system. Let the normal to the inter-face planes be the x-axis, then the interface planes are parallel to the (y,z)-plane and the multilayerstack (ni, di) is defined as follows:

n(x) = n0, x < x0,n1, x0 < x < x1,n2, x1 < x < x2,...nN , xN−1 < x,

(3.30)

where ni is the complex refractive index of layer i and where xi marks the position of the planarinterface between the layers i and (i+ 1). The layers have a thickness di equal to xi+1 − xi.

We now define the z-axis. The normal to the interface planes and the wave vector of the incidentplane wave define a plane, which is called the incident plane. Let the z-axis be the coordinateaxis in this plane orthogonal to the x-axis. As a result, the y-axis is defined and the wave vectorlies entirely in the (x,z)-plane: its y-component is zero, ky= 0. The multilayer structure in thiscoordinate system is depicted in Figure 3.7 (a).

As we noted before, it is sufficient to treat the problem for two orthogonal polarizations. Theeasiest way is to solve the problem once for the TE-polarization (s-wave) with the electric fieldparallel to the interfaces of the stack and once for the TM-polarization (p-wave) with the magnetic

3–10

Figure 3.7: A multilayer stack of dielectric media.

field parallel to the interfaces of the stack. The orientation of both polarizations is illustrated inFigure 3.7 (b). In the following calculations, the two problems - one for each polarization - arediscriminated by the polarization dependent reflection and transmission coefficients. When weconsider only one polarization (TE or TM), the electric field is fully described by its amplitudeE(x, y, z). In each layer, all contributions to the forward propagating wave (towards increasingx) have the same direction and thus form one plane wave. The same holds for the backwardpropagating wave (towards decreasing x). The total field in layer i can thus be described as

E(x, y, z) = AF e−j(kx,ix+kzz) +ABe

−j(−kx,ix+kzz)

= AF e−jkx,ixe−jkzz +ABe

+jkx,ixe−jkzz

= EF (x)e−jkzz + EB(x)e−jkzz(3.31)

where kx,i is the x-component of the wave vector ki in layer i:

kx,i =[(nik0)2 − k2

z

]1/2, i = 0, 1, 2, ..., N, (3.32)

where k0 = (ω/c). Due to boundary conditions that relate the field amplitudes of the incident,reflected and transmitted wave at the interfaces, kz remains constant. When working with losslessmedia and in the absence of total internal reflection, kx,i can be related to the ray angle θi (seeFigure 3.8) in the following way:

kx,i = nik0 cos θi. (3.33)

However, it should be noted that kx,i is a complex number and that its imaginary part can benonzero. This will be the case when =(ni) is nonzero or when kz > nik0. The former correspondsto wave propagation in a lossy medium, the latter corresponds to incident angles that are biggerthan the critical angle for total internal reflection. When total internal reflection occurs, the electricfield vector decreases exponentially in the x-direction and the attenuation occurs within a distanceof q−1 where (if we choose +j as the solution for

√−1):

qi =[k2z − nik2

0

]1/2, i = 0, 1, 2, ..., N, (3.34)

3–11

Figure 3.8: (a) Reflection and transmission at an interface. (b) Translation through a layer with thickness diand refractive index ni.

In that case, the plane wave 3.31 is an evanescent wave propagating parallel to the interface sur-faces in the z-direction.

For a given z, the complete transfer matrix T for wave propagation relates the complex amplitudesEF and EB just before and just behind the first and the last interface of the multilayer stack.

[EF (x−0 )EB(x−0 )

]= T0N

[EF (x+

N−1)EB(x+

N−1)

]=[T 0N

11 T 0N12

T 0N21 T 0N

22

] [EF (x+

N−1)EB(x+

N−1)

]. (3.35)

We first build the individual transfer matrices for wave propagation through an interface andfor wave propagation through a layer and will then calculate the complete transfer matrix bymultiplying all individual ones. We start with the transfer matrix for wave propagation throughan interface. Consider the interface between layer i and layer j (Figure 3.8 (a)). For a given z,the amplitudes of the forward and backward propagating waves just before and just behind theinterface are related in the following way:

[EF (x+

i )EB(x−i )

]=[tij rjirij tji

] [EF (x−i )EB(x+

i )

](3.36)

where we use the complex reflection and transmission coefficients (Fresnel coefficients) for thesituation where only one wave is incident on the interface. This is actually the scattering matrixdescription of the interface between two layers.

In the case of TE-polarization, the Fresnel coefficients are given by:

rij =EB(x−i )EF (x−i )

=kix − k

jx

kix + kjx(3.37)

tij =EF (x+

i )EF (x−i )

= 1 + rij =2kix

kix + kjx(3.38)

3–12

and in the case of TM-polarization:

rij =EB(x−i )EF (x−i )

=n2i kjx − n2

jkix

n2i kjx + n2

jkix

(3.39)

tij =EF (x+

i )EF (x−i )

=ninj

(1 + rij) (3.40)

or their equivalent as a function of the refractive indices ni, nj and ray angles θi, θj , for TE-polarization:

rij =EB(x−i )EF (x−i )

=ni cos θi − nj cos θjni cos θi + nj cos θj

(3.41)

tij =EF (x+

i )EF (x−i )

= 1 + rij =2ni cos θi

ni cos θi + nj cos θj(3.42)

and for TM-polarization:

rij =EB(x−i )EF (x−i )

=nj cos θi − ni cos θjnj cos θi + ni cos θj

(3.43)

tij =EF (x+

i )EF (x−i )

=ninj

(1 + rij) =2ni cos θi

nj cos θj + ni cos θj(3.44)

For perpendicular incidence, there is no difference between the TE and TM case. Equation (3.37)differs however from equation (3.39) in this case. This is caused by the different definition of thedirection of the unit vectors for theE-field, in the TE and TM case. Figure 3.9 depicts the definitionof the unit vectors for the E and H-field for the incident, reflected and refracted wave.

Equations (3.36) can be rewritten in such a way that a relationship is described between the for-ward and backward propagating wave in layer i and the forward and backward propagatingwave in layer j:

[EF (x−i )EB(x−i )

]=

[1tij

− rjitij

rijtij

tji − rijtijrji

] [EF (x+

i )EB(x+

i )

]. (3.45)

We make use of the symmetry relations of the Fresnel coefficients

rij = −rji (3.46)tijtji − rijrji = 1 (3.47)

to simplify this expression and get:

[EF (x−i )EB(x−i )

]=

1tij

[1 rijrij 1

] [EF (x+

i )EB(x+

i )

]. (3.48)

3–13

Figure 3.9: Magnitude and phase of the reflection coefficient in function of incidence angle for (a) externalreflection (nj/ni = 1.5) and TE polarization, (b) external reflection (nj/ni = 1.5) and TM polarization,(c) internal reflection (ni/nj = 1.5) and TE polarization and (d) internal reflection (ni/nj = 1.5) and TMpolarization.

3–14

Figure 3.10: Inspection of Equation (3.52). The phase change Φ corresponds to the path length |AB′| andnot to the path length |AC|.

Thus, the transfer matrix Tij for wave propagation through the interface between layers i and jreads:

Tij =1tij

[1 rijrij 1

]. (3.49)

The next step is to build the transfer matrix for wave propagation through a layer. Consider thelayer i. (Figure 3.8 (b)). For a given z, the amplitudes of the forward and backward propagatingwaves just before and just behind the interfaces of the adjacent layers are related in the followingway:

EF (x−i ) = EF (x+i−1)e−jkx,idi

EB(x+i−1) = EB(x−i )e−jkx,idi ,

(3.50)

with di the thickness of layer i. This follows straightforwardly from Equations (3.31). Thus, thetransfer matrix Ti for wave propagation through layer i reads:

Ti =[ejΦi 0

0 e−jΦi

](3.51)

in which we defined Φi = kx,idi, which in general is a complex quantity. In a lossless medium andin the absence of total internal reflection, Φi is a real quantity:

Φi = kx,idi =2πλ0nidi cos θi, (3.52)

where kx,i was substituted using (3.33). Equation (3.52) might look surprising at first sight. Clearly,Φi represents a phase change but the phase change does not correspond to the path length |AC|.Careful inspection of Equation (3.52) shows that the phase change Φ corresponds to the pathlength |AB′|. The reason for this is that in the transfer matrix formalism, the transfer matricesalways describe relationships between EF and EB at a constant z-level. Thus Φ corresponds tothe distance between the two phase fronts through points A and B, equal to |AB′|.

3–15

The complete transfer matrix for the wave propagation through the layered medium can now befound by multiplication of all individual transfer matrices.

[EF (x−0 )EB(x−0 )

]= T0N

[EF (x+

N−1)EB(x+

N−1)

]=[T 0N

11 T 0N12

T 0N21 T 0N

22

] [EF (x+

N−1)EB(x+

N−1)

](3.53)

where

T0N = T01T1T12T2...T(N−1)T(N−1)N . (3.54)

Equation (3.53) is known as the matrix formulation for wave propagation through multilayer sys-tems. Now, this expression can be used for solving a variety of wave propagation problems. Letus first apply the transfer matrix method to our original problem: a plane wave is incident on astack of layers. First calculate the individual transfer matrices according to the described method(equations (3.49) and (3.51)). Next, calculate the complete transfer matrix by multiplication usingEquation (3.53). As there is no incident field from the right side, we have EB(x+

N−1) = 0. So, weare left with the following matrix formulation for wave propagation in the multilayer stack:

[EF (x−0 )EB(x−0 )

]=[T 0N

11 T 0N12

T 0N21 T 0N

22

] [EF (x+

N−1)0

]. (3.55)

This leads to two equations and two unknowns, namely EB(x−0 ) and EF (x+N−1), which can now

easily be solved for. The reflection and transmission coefficients follow from the ratiosEB(x−0 )/EF (x−0 )and EF (x+

N−1)/EF (x−0 ) respectively.

r = EB(x−0 )

EF (x−0 )= T 0N

21

T 0N11

t =EF (x+

N−1)

EF (x−0 )= 1

T 0N11

(3.56)

From the amplitude reflection and transmission coefficients, the power reflection and transmissioncoefficients can be easily calculated. The power density of a (forward or backward) plane wave inlayer i is given by the real part of the Poynting vector:

〈P (x)〉 = Re(S(x))= Re(E(x)×H(x))= Re(n) |E(x)|2

2Z0

(3.57)

with Z0 the impedance of vacuum (=√µ0/ε0= 377Ω). This power expresses the power per unit

area perpendicular to the propagation direction (which differs from layer to layer). In order tocalculate the power reflection and transmission of the multilayer thin film, it is better to expressthe power density per unit of area parallel to the layers. This modified power density is given by:

〈P (xi)||〉 = cos θi〈P (xi)〉= Re(n) cos θi

|E(xi)|22Z0

(3.58)

3–16

This allows to calculate the power reflection and transmission coefficients of the thin film:

R = Re(n0) cos θ0 |EB(x0)|2Re(n0) cos θ0 |EF (x0)|2 = |r|2

T = Re(nN ) cos θN |EF (xN )|2Re(n0) cos θ0 |EF (x0)|2 = Re(nN ) cos θN

Re(n0) cos θ0|t|2 .

(3.59)

To conclude our discussion of the transfer matrix formalism and to show its broad applicability,let us consider a different problem. What if no light is incident, nor from the left nor from theright? This means: EF (x−0 ) = 0 and EB(x+

N−1) = 0. In this case, we are left with the followingmatrix formulation for wave propagation in the multilayer stack:

[0

EB(x−0 )

]=[T 0N

11 T 0N12

T 0N21 T 0N

22

] [EF (x+

N−1)0

]. (3.60)

So, even if no light is incident, this system has nonzero solutions provided that the followingcondition is fulfilled:

T 0N11 = 0. (3.61)

Let us briefly examine this condition. T 0N11 is a function of kz , ω, indices ni and thicknesses ti of the

layers. Given a multilayer structure (ni, di) and a frequency ω, Equation (3.61) can thus be usedto solve for the propagation constant kz of all the confined modes supported by the multilayerstructure. This example is an illustration of the application of the transfer matrix formalism towaveguide problems. We will come back to the transfer matrix method in a later chapter where itwill be used to analyze slab waveguides.

3.4 Applications of thin films

In this section we discuss two important applications of thin film structures: the Fabry-Perot etalonand optical coatings.

3.4.1 Fabry-Perot etalon

The Fabry-Perot interferometer, or etalon, can be considered as the simplest type of optical res-onator. Normally, such an instrument consists of two parallel dielectric mirrors separated at adistance l. Here, we consider a simple structure that consists of a plane-parallel plate of thicknessl and refractive index n immersed in a medium of index n′ (Figure 3.11). This is a simple multi-layer stack and the transfer matrix formalism can be applied in a straightforward way. 3 transfermatrices need to be calculated and multiplied in order to get the complete transfer matrix of the3-layer system. First, we calculate the transfer matrices associated with the two interfaces:

T12 =1t12

[1 r12

r12 1

]=

1t12

[1 rr 1

]. (3.62)

3–17

and

T23 =1t23

[1 r23

r23 1

]=

1t23

[1 −r−r 1

]. (3.63)

where r = r12 = −r23. The transfer matrix associated to the layer between the two interfaces isthe following:

T2 =[ejΦ 00 e−jΦ

]= ejΦ

[1 00 e−2jΦ

](3.64)

where Φ = 2πλ0nl cos θ. Multiplying yields the complete transfer matrix:

T13 = T = T12T2T23 =1

t12t23ejΦ

[1− r2e−2jΦ −r(1− e−2jΦ)r(1− e−2jΦ) −r2 + e−2jΦ

]. (3.65)

When such a Fabry-Perot etalon is illuminated by a light beam, the situation corresponds to oneincident wave from the left of the structure. Thus, the reflection and the transmission of the struc-ture are described by the formulas in Equation (3.55). Substitution and simplification yields forthe reflection and transmission coefficients rFP and tFP according to 3.56:

rFP = T21T11

= r(1−e−2jΦ)1−r2e−2jΦ

tFP = 1T11

= t12t23e−jΦ

1−r2e−2jΦ ,(3.66)

and for the power reflection and transmission coefficients RFP and TFP according to 3.59:

RFP =∣∣∣T21T11

∣∣∣2 = 4r2 sin2 Φ(1−r2)2+4r2 sin2 Φ

TFP =∣∣∣ 1T11

∣∣∣2 = |t12t23|2(1−r2)2+4r2 sin2 Φ

,(3.67)

describing the reflection and transmission of a Fabry-Perot etalon. Note that our basic model ofthe Fabry-Perot etalon does not contain any loss mechanisms, so conservation of energy requiresRFP + TFP = 1, which is indeed the case. In what follows, we will work with the fraction of theintensity reflected R and the fraction of the intensity transmitted T at one interface and refer to itas the mirror’s reflection and transmittance:

R = r212 = r2

23 = r2

T = t12t23.(3.68)

For the moment, we consider no losses, soR+T = 1. Let us now discuss the results of our calcula-tions given by 3.67. First of all, the formula for the transmitted intensity (3.67) looks very similar tothe formula (3.25) we found in section 3.2.4 describing the interference pattern of an infinite num-ber of plane waves with progressively declining amplitude and equal phase difference. In fact,they are the same. This is not surprising at all. It is clear from Figure 3.11 that the transmission ofa Fabry-Perot etalon is resulting from an infinite number of waves interfering with each other. Ifthe incident intensity is taken as unity, the first transmitted contribution has intensity |t12t23|. Thesecond contribution is decreased by a factor |r12r23|, the third contribution again, and so on. Thephase difference between each contribution remains constant and equal to δ = 2Φ = 22π

λ0nl cos θ.

3–18

Figure 3.11: Reflection and transmission at a parallel plate structure such as the Fabry-Perot etalon.

Thus, from the viewpoint of interference, we end up with exactly the formula 3.25 for the tranmis-sion through this structure. The transfer matrix formalism for wave propagation gives the sameresult, as it should be.

Let us now take a closer look at the transmission characteristics of a Fabry-Perot etalon. Accordingto Equation (3.67), transmission is maximal and equal to unity when

Φ =2πnλl cos θ = mπ,m = integer, (3.69)

where θ is the ray angle in the medium. Φ is called the resonator round trip phase. On the otherhand, the transmission is minimal and equal to

TFP,min =(1−R)2

(1−R)2 + 4R=

11 + 4R

(1−R)2

(3.70)

when the resonator round trip phase equals

Φ =2πnλl cos θ = (2m− 1)

π

2,m = integer. (3.71)

By using λ = c/ν, the condition for maximal transmission can also be written as

νm = mc

2nl cos θ,m = integer, (3.72)

where ν is the optical frequency. For a given l and n, Equation (3.72) defines the resonance frequen-cies of the Fabry-Perot etalon. At resonance, all transmitted contributions interfere constructively.Off resonance, the contributions do no longer interfere constructively. Two neighboring resonancefrequencies are separated by the so-called free spectral range:

∆ν = νm+1 − νm =c

2nl cos θ,m = integer. (3.73)

At this point, it is interesting to look back at Figure 3.5 depicting the intensity of an infinite numberof interfering waves with declining amplitude and equal phase difference. It should be clear nowthat this figure also depicts the transmission of a Fabry-Perot cavity as a function of phase, see

3–19

Fig. 3.12. However, note that the Fabry-Perot peaks are at the positions Φ = mπ, which correspondto δ = 2mπ in Fig. 3.5.

The finesse F is directly related to the mirror’s reflection R at each interface by

F =π√R

1−R. (3.74)

Thus, the higher the mirror’s reflectionR at each interface, the higher is the finesse and the sharperare the transmission peaks:

R = 50% → F ∼= 4R = 90% → F = 30R = 98% → F = 156

(3.75)

Another important dimensionless quantity in connection with resonances is the quality factor orQ-factor. It is defined as

Q =ωr∆ω

(3.76)

with ωr the (angular) resonant frequency, and ∆ω the FWHM-bandwidth (Full Width at Half Max-imum). It is a measure of the sharpness of the resonance in the spectrum. We see in Fig. 3.12 e.g.that a Fabry-Perot with strongly reflecting mirrors gives sharper peaks with a larger Q, intuitivelyindicating that the resonance is ‘better defined’ or ‘purer’.

Now we investigate the phase characteristics of the Fabry-Perot etalon. If we write the amplitudetransmission coefficient 3.66 as

t = |t| e−j∠t, (3.77)

and examine the phase shift ∠t as a function of the cavity length l (or equivalently Φ), we note thata strong dispersion of the phases versus Φ exists at resonance. This is shown in Figure 3.12. Notethat there is a link to the concept of group delay, defined as

d∠tdω

(3.78)

which measures the propagation time of a signal through the structure. We clearly note an increaseof the group delay at resonance, which corresponds to the intuition that light then spends a longertime in the cavity. Remark that without the cavity structure, the phase would be a straight line(∠t = k0nd) in Figure 3.12.

When designing a Fabry-Perot resonator with high finesse, one needs to ensure that TFP,min givenby Equation (3.70) is as small as possible. Therefore R needs to be close to 1. In practice, thisis difficult because of the available materials. Indeed, the refractive index of optical materials islimited to about n ≈ 4, which means for a Fabry-Perot in air a maximal reflection of onlyR ≈ 36%.

One solution to this problem is offered by applying metal coatings on the interfaces. Due to thepresence of such a metal coating, an additional phase difference will occur upon reflection. Soassume:

r =√Re+jα. (3.79)

With this r one can calculate for the transmission

TFP =∣∣∣ T

1−Re−j2(φ−α)

∣∣∣2 (3.80)

3–20

Figure 3.12: Power transmission/reflection and phase of transmission for a Fabry-Perot etalon as a functionof cavity length l (or equivalently Φ). Solid line: R = 30%, dashed line: R = 80%.

where T is no longer equal to 1−R but to 1−R− A with A the intensity absorption in the metalfilm. As a result, the maximal transmission drops from unity to

TFP,max =T 2

(1−R)2=[1− A

1−R

]2

. (3.81)

The relation between the ratio TFP,max

TFP,minand the mirror’s reflection R is depicted in Figure 3.13.

To provide more insight into the operation of the Fabry-Perot cavities we plot the energy in andaround the cavities in various situations, see Figure 3.14. One notices that at resonance the energyin the cavities is enhanced, and the transmission is unity. Off resonance (middle graphs) we finda standing wave in reflection, and a small field in the cavity and also a small transmission.

Most Fabry-Perot etalons are made of two identical mirrors. In fact, the symmetry of the mirrorreflection is important to obtain high finesse. Any asymmetry in the reflection will lead to a de-crease in either transmission or finesse. (It is left as an exercise to investigate the transmissionproperties of an asymmetric Fabry-Perot etalon in detail.)

3.4.2 Coatings

Multilayer stacks can be used as coatings that alter the optical properties of a substrate, for exam-ple increasing or decreasing reflection. Anti-reflective coatings and highly-reflective coatings aretwo examples that are widely used.

3–21

Figure 3.13: TF P,maxTF P,min

as a function of R.

AR-coatings: quarter-wave layer

Minimal reflection or equivalently maximal transmission is desirable in many applications. Oneexample is efficient coupling of light from a fibre into a waveguide. In designing an anti-reflective-coating one needs to ensure that the reflection at the front of the coating interferes destructivelywith the reflection at the back of the coating. Assume a film thickness d. If n1 < n2 < n3 with n2

the refractive index of the coating, the two waves will interfere destructively if

d =14λ0

n2, (3.82)

hence the name quarter-wave layer. The question remains which refractive index is needed forthe coating. From the transfer matrix formalism for this 3-layer structure, one easily obtains

r =r12 + r23e

−2jΦ

1 + r12r23e−2jΦ=

r12 − r23

1− r12r23, (3.83)

where we used Φ = 2πλ0n2d = π

2 . Given that rij = ni−njni+nj

, the reflection will be zero when

n2 =√n1n3. (3.84)

In practice however, materials with such a refractive index may not exist. Nevertheless, usingavailable materials with an index of refraction close to that given by Equation (3.84), a great re-duction in reflection is obtained. We also notice that by using materials with an index of refractionbetween n1 and n3, a single-layer coating will always reduce the reflection, regardless of the layerthickness. Thus, the reflection of a coated surface is always lower than that of an uncoated one,

3–22

Figure 3.14: Energy (ε|E|2) in the Fabry-Perot cavities for R = 30% (left) and R = 80% (right). The situationis shown for Φ/π = 2 (top), Φ/π = 2.5 (middle) and Φ/π = 3 (bottom). The cavity is located between z = 4dand z = 5d. Light is incident from the left with |E|2 = 1.

3–23

Figure 3.15: Reflection of a HR-coating for a He-Ne laser at wavelength λ0 =633nm. n1 =2.32 (ZnS), n2=1.38(MgF2). A reflection of 98.9 % is achieved already after 13 layers.

provided that the index of refraction of the coating is between that of the two media. In somesense, the coating smooths dielectric discontinuity.

High-reflective coatings

High-reflective mirrors are desirable in many applications. These include high-finesse Fabry-Perotinterferometers and low-loss laser resonators. Mirrors made of metallic films such as silver, alu-minum or gold are generally of high reflection. For example, a silver mirror can achieve reflectionapproaching 99 % in the visible spectrum. Approximately 1 % of light energy penetrates the sur-face of the metal and gets absorbed in the bulk of the metal. These metallic mirrors cannot be usedwith high-power lasers because even a small fraction of absorption can cause severe heating prob-lems. Thus there is a need to design high-reflection mirrors by using materials that have (almost)no absorption.

The dielectric layered structure that consists of alternating quarter-wave layers of two differentmaterials is the simplest way to obtain high reflection. This is the so-called Bragg reflector. If for acertain wavelength λ0, the thicknesses d1, d2 and the refractive indices n1 and n2 of the consecutivelayers can be controlled so that:

n1d1 = n2d2 =λ0

4, (3.85)

then the reflected beams from the different interfaces will all interfere constructively, leading to apeak in the reflection spectrum for this wavelength.

Using the matrix method, the peak reflection RHR,max can be calculated and is given by

RHR,max =

1−(nsna

)(n2n1

)2N

1 +(nsna

)(n2n1

)2N

2

, (3.86)

where ns is the refractive index of the substrate, na that of air and N is the number of periods.RHR,max converges to 1 as N increases. The convergence improves as the ratio n1

n2becomes larger.

This is illustrated in Figure 3.15 for the case of a high reflective coating for a He-Ne laser at wave-length λ0 = 633nm and using a quarter-waves stack of ZnS (n1 =2.32) and MgF (n2=1.38). How-ever, Bragg reflectors can provide high reflection over any desired spectral regime of interest by

3–24

Figure 3.16: Reflection spectrum of Bragg coatings with different index contrasts ∆n and number of layersN : (a) ∆n = 0.5, (b) ∆n = 2.5. Solid (dashed) curves correspond with N = 3 (N = 10), respectively.

properly tailoring the layer thicknesses. It can be proven that the bandwidth of these reflectors isgiven by

∆λλ

=4π

sin−1 |n2 − n1|n2 + n1

, (3.87)

which can be approximately written as

∆λλ

=2π

∆nn, (3.88)

provided that n1∼= n2

∼= n. Thus, a high bandwidth is obtained by increasing the differencebetween the refractive indices of the layers. These trends are illustrated in Figure 3.16.

To obtain more insight in the Bragg reflector we plot a few field profiles in Figure 3.17. In thebandgap we expect a strong reflection, leading to an exponentially decaying field in the reflector,which is confirmed in Fig. 3.17(d). On the sides of the bandgap there are frequencies with reflec-tion equal to zero (and thus total transmission), these correspond to Fabry-Perot like resonances,built up in between the ends of the Bragg reflector. Indeed, the field profile in Fig. 3.17(c) showsa bump in the profile in the reflector, and total transmission. In addition, the next reflection min-imum is shown in Fig. 3.17(a), which indicates a higher order resonance, as it has two maxima inthe field profile. In Fig. 3.17(b) the situation in between the previous resonances is plotted, at thepoint where the reflection has a (relative) maximum. We notice an intermediate situation, withone and a half lobes inside the cavity, leading to a significant reflection (indicated by the standingwave pattern).

Bragg reflectors can be made to reflect broad bands of light by stacking up several periodic layeredmedia with different periods. In this case, each reflector acts as a band rejection filter for eachwavelength. If the bandwidths are wide enough to have substantial overlap, the whole structurecan reject a broad band of light.

3–25

Figure 3.17: Field plots in and around a Bragg reflector. Light with |E| = 1 is incident from the left side.The reflector, located between z/period = 5 and 10, has 10 periods with n1 = 1, n2 = 1.5.

3–26

Figure 3.18: Schematic of a Fabry-Perot cavity with Bragg reflector mirrors.

Figure 3.19: Reflection spectra of Bragg structures with n1 = 1, n2 = 1.5. (a) Solid line: n1-defect in centerwith lcen = λ0/2n1, 5 periods on left and right. Dashed line: Bragg without defect and 10 periods. (b)Thicker n1-defect in center with lcen = 5.5λ0/2n1, 5 periods on left and right.

Fabry-Perot with Bragg reflector

One can combine the concepts of the Fabry-Perot cavity with the Bragg reflector to create a veryhigh quality resonance. This can be done by enlarging one of the layers in the middle of a Braggreflector. In this way the sides of the Bragg reflector act as good mirrors (in the bandgap), and the‘defect’ layer accommodates a Fabry-Perot type cavity mode. Thus, ideally the Bragg wavelengthshould correspond with the Fabry-Perot resonance wavelength.

A sketch of such a structure is shown in Fig. 3.18, with lcen the thickness of the defect layer. Re-flection spectra are depicted in Fig. 3.19. We see a sharp dip in reflection in the center of thebandgap in Fig. 3.19(a)(solid). The spectrum of the Bragg reflector without the defect is indi-cated with a dashed line. For this particular defect mode we chose lcen = λ0/2n1, thus givinga fundamental Fabry-Perot resonance at λ0, which is also the center of our Bragg bandgap (forl1 = λ0/4n1, l2 = λ0/4n2). A thicker defect layer can give rise to multiple Fabry-Perot type modesin one bandgap, this is illustrated in Fig 3.19(b).

3–27

Bibliography

[Bae06] R. Baets. Photonics (Syllabus). Universiteit Gent, 2006.

[Yeh88] P. Yeh. Optical Waves in Layered Media. John Wiley and Sons, 1988.

3–28